Differential Eqns Separable Equation?

[Prologue]

Homework Statement

Problem given basically in the beginning of the book, Sec. 1.4, about separable equations.

Find general solutions (implicit if necessary, explicit if convenient) of the diff. eq.

[tex]dy/dx+2xy^2=0[/tex]

Homework Equations

There was a previous section where they popped this out of thin air:
A solution of [tex]dy/dx=y^2[/tex] is [tex]y=1/c-x[/tex]

I'm presuming that that is supposed to be helpful somehow.

The Attempt at a Solution

[tex]dy/dx=-2xy^2 => dy/y^2=-2xdx[/tex]

Now I can't integrate the left side.


I'm sure that is an abysmal way of going at it. So any help would be appreciated.

 

[HallsofIvy]

Homework Statement

Problem given basically in the beginning of the book, Sec. 1.4, about separable equations.

Find general solutions (implicit if necessary, explicit if convenient) of the diff. eq.

[tex]dy/dx+2xy^2=0[/tex]

Homework Equations

There was a previous section where they popped this out of thin air: [tex]A solution of dy/dx=y^2 is 1/c-x[/tex]

I'm presuming that that is supposed to be helpful somehow.

The Attempt at a Solution

[tex]dy/dx=-2xy^2 => dy/y^2=-2xdx[/tex]

Now I can't integrate the left side.


I'm sure that is an abysmal way of going at it. So any help would be appreciated.

That's a perfectly good way of going at it. The question is why are you having trouble integrating y^-2dy? Didn't you learn a general formula for integrating powers? (Other than -1)

 

[Prologue]

I haven't learned a formula for that. I will bet that it involves grabbing the derivative of the denominator from somewhere and thinking of it as a u-substitution but I'm not seeing where that is possible.


EDIT: Ok I just ignored that y was a function of x and integrated, and it worked out perfectly fine. I am supposing that just this idea is a point of this exercise. When you integrate something in this way, you have an equation and integrate both sides. When you do it that way y is no longer taken as being a function of x. I can accept that as the point, but now why is that possible?

 

[konthelion]

Ah, I think you're looking at it like this [tex]\frac{f'(x)}{[f(x)]^2}[/tex] in your head, and somehow you think of it as a u-substitution where u = f(x).

Basically the general formula for integrating is

[tex]\int{x^rdx}= \frac{1}{r+1} x^{r+1}+C[/tex] where [tex]r \neq -1[/tex]

 

[Prologue]

Thanks for the reply, konthelion. I was in the process of editing the last post when you posted, I apologize.

I do know that general rule of integration (it has been severely beaten into my brain) but I was having trouble with the y. My head is saying 'Hey, hold on a minute y is a function of x, you can't do that.' Apparently my head is wrong.